Question

Perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. In Cleveland, a random sample of 73 mail carriers showed that 10 had been bitten by an animal during one week. In Philadelphia, in a random sample of 80 mail carriers, 16 had received animal bites. Is there a significant difference in the proportions? Use \(\alpha=0.05 .\) Find the \(95 \%\) confidence interval for the difference of the two proportions.

Short answer

There is no significant difference in animal bite proportions between Cleveland and Philadelphia mail carriers.

Step-by-step solution

Stating Hypotheses and Identifying the Claim

First, we'll set up our null hypothesis \(H_0\) and alternative hypothesis \(H_1\). The null hypothesis is that there is no significant difference between the proportions of mail carriers bitten in Cleveland and Philadelphia. Mathematically, this is expressed as \(H_0: p_1 = p_2\). The alternative hypothesis, which is the claim, is that there is a significant difference: \(H_1: p_1 eq p_2\).

Finding the Critical Value(s)

For a two-tailed test at \(\alpha = 0.05\), we find the critical values using the standard normal (Z) distribution. The critical values for \(\alpha = 0.05\) are \(Z = \pm 1.96\). This means if the test statistic falls beyond these values, we reject the null hypothesis.

Computing the Test Value

First, calculate the sample proportions: \(\hat{p}_1 = \frac{10}{73}\) for Cleveland and \(\hat{p}_2 = \frac{16}{80}\) for Philadelphia. Therefore, \(\hat{p}_1 = 0.137\) and \(\hat{p}_2 = 0.200\). The pooled proportion is \(\hat{p} = \frac{10 + 16}{73 + 80} = 0.170\). The test statistic \(Z\) is calculated using:\[Z = \frac{(\hat{p}_1 - \hat{p}_2)}{\sqrt{\hat{p} (1-\hat{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} = \frac{0.137 - 0.200}{\sqrt{0.170(1-0.170)\left(\frac{1}{73} + \frac{1}{80}\right)}} = -1.052\]

Making the Decision

Compare the test statistic \(Z = -1.052\) to the critical values \(Z = -1.96\) and \(Z = 1.96\). Since \(Z = -1.052\) falls within the range \(-1.96\) to \(1.96\), we fail to reject the null hypothesis. This means there is not enough evidence to claim a significant difference in proportions.

Summarizing the Results

Based on our hypothesis test, we conclude that there is no significant difference between the proportions of mail carriers bitten by animals in Cleveland and Philadelphia during the observed week. To compute the 95% confidence interval for the difference in proportions, use:\[(\hat{p}_1 - \hat{p}_2) \pm Z \times \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} = -0.063 \pm 1.96 \times 0.071\]The 95% confidence interval is \(-0.063 \pm 0.139\), or \(-0.202, 0.076\). This interval contains zero, supporting our conclusion of no significant difference.

Key concepts

Critical Values in Hypothesis Testing

Critical values are thresholds that help us decide the outcome of hypothesis tests. For a two-tailed test, like in the example provided, critical values determine the cutoff points on a distribution curve. These points mark the boundaries for accepting or rejecting the null hypothesis.

In our case, with a significance level (b1) of 0.05, we find the critical values from the standard normal distribution (Z-distribution). For b1 at 0.05, critical values are b1= b11.96. This implies that if our test statistic exceeds these critical limits, we reject the null hypothesis. Otherwise, we do not reject it.

Finding critical values is essential because they provide a clear criterion for making decisions in hypothesis testing. They act as the lines we look beyond to infer about our hypotheses.

Confidence Intervals: Understanding Estimate Range

Confidence intervals (CIs) give a range in which we expect the true value of the population parameter to fall, with a certain level of confidence, usually 95%.

In this exercise, we computed the 95% confidence interval for the difference in proportions of mail carriers bitten by animals between the two cities. The formula we used incorporates the difference in sample proportions and helps assess the reliability of this estimation. A 95% confidence interval suggests that if we repeated our sampling process multiple times, 95% of the computed intervals would contain the true difference in proportions.
The interval calculated was b9-0.202, 0.076b9. This range contains zero, supporting our hypothesis test conclusion that there's no significant difference in the actual proportions being compared. This means we are reasonably confident that any observed difference may simply be due to random chance.

Null Hypothesis: The Default Assumption

The null hypothesis (b9H_0b9) is a default statement proposing that no effect or difference exists in the context of an investigation. It acts as a starting point for hypothesis testing.

In the given exercise, the null hypothesis posited that there's no significant difference between mail carrier bite proportions in Cleveland and Philadelphia. Thus, mathematically, it was expressed as b9H_0: p_1 = p_2b9 where b9p_1b9 and b9p_2b9 denote the proportions of bitten carriers in the respective cities.

If the evidence from our sample is insufficient to reject b9H_0b9 (as it turns out here), we maintain the position that there is no notable difference.

• The null hypothesis provides a baseline for comparison.
• It's essential to validate findings with statistical support.

This concept fundamentally guides the direction of hypothesis testing.